Calculate the time it takes light to travel a significant distance such as between objects in the Solar System

6.1 – The Solar System and Light‑Travel Times

Learning objectives

• Identify the main components of the Solar System (Sun, planets, dwarf planets, moons, asteroids, comets).
• State the order of the eight planets from the Sun.
• Explain the astronomical unit (AU) and convert between AU, kilometres and metres.
• Calculate the average orbital speed of a planet using v = 2πr / T.
• Derive the light‑travel time between any two objects in the Solar System using t = d / c.
• Link the distances involved to gravitational force, Kepler’s third law and spacecraft communication.
• Recall the basic Earth‑system facts needed for the “Earth” sub‑topic (rotation, tilt, lunar orbit).

6.1.1 – The Earth (brief recap)

• Rotation: 1 rotation ≈ 24.0 h (86 400 s) → day/night cycle.
• Axial tilt: ≈ 23.5° → causes the seasons.
• Orbit around the Sun: average distance 1 AU = 1.496 × 10⁸ km; orbital period ≈ 365.25 days (3.156 × 10⁷ s).
• Moon: orbital period ≈ 27.3 days; average distance ≈ 3.84 × 10⁵ km.

6.1.2 – Overview of the Solar System

The Solar System consists of:

• The Sun – a G‑type main‑sequence star containing > 99 % of the system’s mass.
• Eight planets (in order from the Sun):
  

  Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune.

• Dwarf planets – e.g. Pluto, Eris, Haumea, Makemake, Ceres.
• Moons – natural satellites (e.g. Earth’s Moon, Jupiter’s Galilean moons).
• Small bodies – asteroids (main belt, near‑Earth), comets (Kuiper Belt, Oort Cloud), meteoroids.

6.1.3 – The Astronomical Unit (AU)

• 1 AU is defined as the mean Sun‑Earth distance.
• Numerically, 1 AU = 1.496 × 10⁸ km = 1.496 × 10¹¹ m.
• Using AU lets us compare planetary distances without handling very large numbers (e.g. “Mars is 1.52 AU from the Sun”).

6.1.4 – Orbital speed of a planet

For a (near‑)circular orbit the average orbital speed is

\[

v=\frac{2\pi r}{T}

\]

where

• r = orbital radius (≈ average distance from the Sun) in metres,
• T = orbital period in seconds.

Worked example – Earth

• r = 1 AU = 1.496 × 10¹¹ m
• T = 1 yr = 3.156 × 10⁷ s (Gregorian year)

\[

v_{\text{Earth}}=\frac{2\pi(1.496\times10^{11})}{3.156\times10^{7}}\approx2.98\times10^{4}\ \text{m s}^{-1}=29.8\ \text{km s}^{-1}

\]

Worked example – Mars

• Average distance = 1.52 AU → r = 1.52 × 1.496 × 10¹¹ m = 2.27 × 10¹¹ m
• Orbital period = 1.88 yr → T = 1.88 × 3.156 × 10⁷ s = 5.94 × 10⁷ s

\[

v_{\text{Mars}}=\frac{2\pi(2.27\times10^{11})}{5.94\times10^{7}}\approx2.41\times10^{4}\ \text{m s}^{-1}=24.1\ \text{km s}^{-1}

\]

6.1.5 – Speed of light

In vacuum, the accepted value is

\[

c = 3.00\times10^{8}\ \text{m s}^{-1}

\]

(kept to three significant figures, as required by the syllabus).

6.1.6 – Light‑travel time

The time for light (or a radio signal) to cover a distance d is

\[

t = \frac{d}{c}

\]

• t = time in seconds (convert to minutes or hours as needed).
• d = distance in metres.
• c = 3.00 × 10⁸ m s⁻¹.

6.1.7 – Typical Solar‑System distances

*Pluto is retained for historical reference; it is classified as a dwarf planet.

6.1.8 – Example calculation – Sun to Earth

1. Distance: d = 1.496 × 10¹¹ m.
2. Apply t = d / c:

   \[

   t = \frac{1.496\times10^{11}}{3.00\times10^{8}} = 4.99\times10^{2}\ \text{s}

   \] (3 SF)

3. Convert to minutes: \(\displaystyle \frac{4.99\times10^{2}}{60}=8.32\ \text{min}\) (2 SF).
4. Result: Light needs ≈ 8.3 minutes to travel 1 AU.

6.1.9 – Step‑by‑step method for any pair of objects

1. Obtain the average distance (usually given in km).
2. Convert to metres: multiply by 10³.
3. Insert the value into \(t = d / c\) with \(c = 3.00\times10^{8}\ \text{m s}^{-1}\).
4. Calculate t in seconds.
5. Convert seconds → minutes (÷ 60) or hours (÷ 3600) as required, keeping the appropriate number of significant figures.

6.1.10 – Connecting ideas (extended)

• Gravitational force: \(F = G\frac{m1m2}{r^{2}}\).

  The same distance r appears in the light‑travel‑time formula; larger r means weaker gravity and longer travel times.

• Deriving orbital speed from gravitation:

  Equating centripetal force \(\frac{mv^{2}}{r}\) to the gravitational attraction \(G\frac{Mm}{r^{2}}\) gives

  \[

  v = \sqrt{\frac{GM}{r}}.

  \]

  Substituting \(v = 2\pi r / T\) leads to Kepler’s third law \(T^{2}\propto r^{3}\). Both relationships explain why objects farther from the Sun have longer orbital periods and longer light‑travel times.

• Spacecraft communication: Radio signals travel at the same speed as light, so mission planners must add the light‑travel delay to command‑and‑control timelines (e.g. ~ 8 min to Mars when it is at opposition).
• Error‑analysis tip (AO2): When converting units or performing the division \(d/c\), keep at least three significant figures in intermediate steps, then round the final answer to the number of figures required by the question (usually 2 SF for light‑travel times).

6.1.11 – Practice questions

1. How long does light take to travel from the Sun to Mars? Give your answer in minutes (one decimal place).
2. Calculate the light‑travel time from Earth to Jupiter. Express your answer in minutes.
3. A signal is sent from Earth to a spacecraft orbiting Saturn. If the distance is \(1.433\times10^{12}\ \text{m}\), how many seconds does the signal require?
4. Compare the light‑travel times from the Sun to Mercury and from the Sun to Neptune. Which is longer and by how many minutes?
5. If a hypothetical planet were located \(2.5\times10^{13}\ \text{m}\) from the Sun, what would be the light‑travel time in hours?

Answers (for teacher use)

1. Sun–Mars: \(d = 2.279\times10^{11}\ \text{m}\) → \(t = 7.60\times10^{2}\ \text{s} = 12.7\ \text{min}\).
2. Earth–Jupiter: distance ≈ 5.20 AU = \(7.785\times10^{11}\ \text{m}\) → \(t = 2.60\times10^{3}\ \text{s} = 43.3\ \text{min}\).
3. Saturn distance: \(t = \dfrac{1.433\times10^{12}}{3.00\times10^{8}} = 4.78\times10^{3}\ \text{s}\) (≈ 4 800 s).
4. Sun–Mercury: \(t = \dfrac{5.79\times10^{10}}{3.00\times10^{8}} = 1.93\times10^{2}\ \text{s}=3.2\ \text{min}\).
   

   Sun–Neptune: \(t = \dfrac{4.503\times10^{12}}{3.00\times10^{8}} = 1.50\times10^{4}\ \text{s}=250\ \text{min}\).
   

   Neptune is longer by ≈ 247 min.

5. Distance \(=2.5\times10^{13}\ \text{m}\) → \(t = \dfrac{2.5\times10^{13}}{3.00\times10^{8}} = 8.33\times10^{4}\ \text{s}\).
   

   Convert to hours: \(\dfrac{8.33\times10^{4}}{3600}=23.1\ \text{h}\).

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